minkowski inequality

**Archi** · January 2nd 2010, 03:00 PM

Let $(X,F,\mu)=([0,a],\beta, \Lambda)$ be the usual Lebesgue measurable space.
Using Holder inequality, prove Minkowski inequality for any $f \in L^2(X),g \in L^2(X)$ , that is $||f+g||_2 \leq ||f||_2 +||g||_2$

i am stuck and dont know what to do. any help would be appreciated.

**putnam120** · January 3rd 2010, 07:20 AM

Imitate the proof of the triangle inequality. (that is start by squaring both sides) Then you can use Holder's inequality with $p=q=2$

**Archi** · January 4th 2010, 03:09 PM

Originally Posted by putnam120

Imitate the proof of the triangle inequality. (that is start by squaring both sides) Then you can use Holder's inequality with $p=q=2$

i am not sure what i should square the both sides of. holders inequality?

**putnam120** · January 4th 2010, 07:23 PM

$\parallel f+g\parallel_2^2=\int (f+g)^2=\int f^2+\int g^2+2\int fg$

$(\parallel f\parallel_2 +\parallel g\parallel_2)^2=\int f^2+\int g^2+2\parallel f\parallel_2\parallel g\parallel_2$

From here you should be able to see how to use Holder's.

Thread: minkowski inequality

LinkBack

Thread Tools

Display

minkowski inequality

Similar Math Help Forum Discussions

Inequality with x in the denominator (rational inequality)

Proving the triangle inequality using the Cauchy-Schwartzs inequality

Help on This Inequality

Minkowski addition-Straight Skeleton Algorithm

Minkowski distance

Search Tags